Theorem sspsstri 4062

Description: Two ways of stating trichotomy with respect to inclusion. (Contributed by NM, 12-Aug-2004.)

Assertion

Ref	Expression
sspsstri	⊢ ((𝐴 ⊆ 𝐵 ∨ 𝐵 ⊆ 𝐴) ↔ (𝐴 ⊊ 𝐵 ∨ 𝐴 = 𝐵 ∨ 𝐵 ⊊ 𝐴))

Proof of Theorem sspsstri

Step	Hyp	Ref	Expression
1		or32 938	. 2 ⊢ (((𝐴 ⊊ 𝐵 ∨ 𝐵 ⊊ 𝐴) ∨ 𝐴 = 𝐵) ↔ ((𝐴 ⊊ 𝐵 ∨ 𝐴 = 𝐵) ∨ 𝐵 ⊊ 𝐴))
2		sspss 4058	. . . 4 ⊢ (𝐴 ⊆ 𝐵 ↔ (𝐴 ⊊ 𝐵 ∨ 𝐴 = 𝐵))
3		sspss 4058	. . . . 5 ⊢ (𝐵 ⊆ 𝐴 ↔ (𝐵 ⊊ 𝐴 ∨ 𝐵 = 𝐴))
4		eqcom 2772	. . . . . 6 ⊢ (𝐵 = 𝐴 ↔ 𝐴 = 𝐵)
5	4	orbi2i 925	. . . . 5 ⊢ ((𝐵 ⊊ 𝐴 ∨ 𝐵 = 𝐴) ↔ (𝐵 ⊊ 𝐴 ∨ 𝐴 = 𝐵))
6	3, 5	bitri 278	. . . 4 ⊢ (𝐵 ⊆ 𝐴 ↔ (𝐵 ⊊ 𝐴 ∨ 𝐴 = 𝐵))
7	2, 6	orbi12i 927	. . 3 ⊢ ((𝐴 ⊆ 𝐵 ∨ 𝐵 ⊆ 𝐴) ↔ ((𝐴 ⊊ 𝐵 ∨ 𝐴 = 𝐵) ∨ (𝐵 ⊊ 𝐴 ∨ 𝐴 = 𝐵)))
8		orordir 942	. . 3 ⊢ (((𝐴 ⊊ 𝐵 ∨ 𝐵 ⊊ 𝐴) ∨ 𝐴 = 𝐵) ↔ ((𝐴 ⊊ 𝐵 ∨ 𝐴 = 𝐵) ∨ (𝐵 ⊊ 𝐴 ∨ 𝐴 = 𝐵)))
9	7, 8	bitr4i 281	. 2 ⊢ ((𝐴 ⊆ 𝐵 ∨ 𝐵 ⊆ 𝐴) ↔ ((𝐴 ⊊ 𝐵 ∨ 𝐵 ⊊ 𝐴) ∨ 𝐴 = 𝐵))
10		df-3or 1102	. 2 ⊢ ((𝐴 ⊊ 𝐵 ∨ 𝐴 = 𝐵 ∨ 𝐵 ⊊ 𝐴) ↔ ((𝐴 ⊊ 𝐵 ∨ 𝐴 = 𝐵) ∨ 𝐵 ⊊ 𝐴))
11	1, 9, 10	3bitr4i 306	1 ⊢ ((𝐴 ⊆ 𝐵 ∨ 𝐵 ⊆ 𝐴) ↔ (𝐴 ⊊ 𝐵 ∨ 𝐴 = 𝐵 ∨ 𝐵 ⊊ 𝐴))

Syntax hints:   ↔ wb 209   ∨ wo 860   ∨ w3o 1100   = wceq 1563   ⊆ wss 3907   ⊊ wpss 3908

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1818  ax-4 1832  ax-5 1933  ax-6 1990  ax-7 2031  ax-9 2155  ax-ext 2737

This theorem depends on definitions:  df-bi 210  df-an 401  df-or 861  df-3or 1102  df-ex 1803  df-cleq 2757  df-ne 2961  df-ss 3924  df-pss 3927

This theorem is referenced by:  ordtri3or  6382  sorpss  7715  sorpssi  7716  funpsstri  36124